PHP/MySqli对照表B ID检查表a ID

我有以下功能，显示数据库中的博客帖子：

function show_blog_posts() { 
    include('connection.php');
    $sql = "SELECT blog.title, blog.content, blog.posted_by, blog.date, article_comments.comments, article_comments.comment_by
                FROM blog LEFT OUTER JOIN article_comments
                ON blog.content_id = article_comments.comment_id
                WHERE blog.content != ''
                ORDER BY blog.content_id DESC";
    $result = mysqli_query($dbCon, $sql);
    while ($row = mysqli_fetch_array($result)) {
        echo 
            "<h5 class='posted_by'>Posted by " . $posted_by = $row['posted_by'] . " on " . $row['date'] . "</h5>" . 
            "<h1 class='content_headers'>" . $title = $row['title'] . "</h1>" . 
            "<article>" . $content = $row['content'] . "</article>" . 
            "<div class='commented_by'>Posted by: " . $row['comment_by'] . "</div>" . 
            "<div class='comments'>Comments: " . $row['comments'] . "</div>";
    }
}

我正在尝试将表A ID与表B ID链接，因此如果ID不匹配，用户将无法发布。如果我的ID不匹配，即使没有文章可以发表，用户仍然可以发表评论

我还编写了以下函数来获取comments表的ID和articles表的ID

function get_article_id($username) { 
    include('db_connection.php');
    $username = sanitize($username);
    $sql = "SELECT content_id FROM `blog` WHERE content_id = '$content_id'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'content_id'));
}

function get_comment_id($comment_id) { 
    include('db_connection.php');
    $comment_id = sanitize($username);
    $sql = "SELECT comment_id FROM `article_comments` WHERE comment_id = '$comment_id'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'comment_id'));
}

如何检测表A id并将其与表B id进行比较？如果

---

另外，我的代码有什么问题，因为sql查询没有执行

首先我想告诉你，一个博客可以有多个评论。因此，您需要小心地与注释表建立关系

article\u comments

表的外键id为

“content\u id”

现在，将您的

show\u blog\u posts

函数的查询更改为类似的无注释关系查询，然后使用另一个查询在

while

循环中获取其注释

这是代码

function show_blog_posts()
{
    include('connection.php');
    $sql = "SELECT blog.title, blog.content, blog.content_id, blog.posted_by, blog.date
            FROM blog
            WHERE blog.content != ''
            ORDER BY blog.content_id DESC";
    $result = mysqli_query($dbCon, $sql);

    if (mysqli_num_rows($result) > 0) {
        while ($row = mysqli_fetch_array($result)) {
            echo
                "<h5 class='posted_by'>Posted by " . $posted_by = $row['posted_by'] . " on " . $row['date'] . "</h5>" .
                "<h1 class='content_headers'>" . $title = $row['title'] . "</h1>" .
                "<article>" . $content = $row['content'] . "</article>";

            $sql = "SELECT article_comments.comments, article_comments.comment_by
            FROM article_comments
            WHERE article_comments.content_id =" . $row['content_id'] . "
            ORDER BY article_comments.comment_id DESC";
            $comments = mysqli_query($dbCon, $sql);

            echo "<div>"; //comments panel
            if (mysqli_num_rows($result) > 0) {
                while ($comment = mysqli_fetch_array($comments)) {
                    echo "<div class='commented_by'>Posted by: " . $comment['comment_by'] . "</div>";
                    echo "<div class='comments'>Comments: " . $comment['comments'] . "</div>";
                }
            }
            echo "</div>"; //comments panel ends
        }
    }

}

function insert_comments($comment_by, $comments, $content_id) {
    include('db_connection.php');
    $sql =  "INSERT INTO article_comments (content_id, comments, comment_by) VALUES ($content_id, '$comments', '$comment_by');";
    mysqli_query($dbCon, $sql);
}

当您在

get\u article\u id

中传递

username

时，让您根据

username

与

content\u id

进行比较

function get_article_id($username) {
    include('db_connection.php');
    $username = sanitize($username);
    //if username is unique
    $sql = "SELECT content_id FROM `blog` WHERE username = '$username'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'content_id'));
}

您必须在

get\u comment\u id

中传递

username

而不是

comment\u id

，因为您要检索

comment\u id

。当您知道注释id时，就不需要调用此函数。我相应地改变了它

function get_comment_id($username) {
    include('db_connection.php');
    $username = sanitize($username);
    $sql = "SELECT comment_id FROM `article_comments` WHERE username = '$username'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'comment_id'));
}

---

INSERT

语句仅适用于INSERT。您不能使用

where

子句在表B上创建引用表a的外键。@Don'tPanic我已尝试创建外键：

ALTER table article\u comments ADD CONSTRAINT comment\u blog\u fk foreign key（blog\u id）references wt.blog（content\u id）on DELETE on UPDATE CASCADE上无操作但我得到了错误#1452-无法添加或更新子行：外键约束失败（
worldtour
#sql-22a4_47b
，约束
注释_blog_fk
外键（
blog_id
）引用
blog
（
content_id
）关于删除，不执行更新级联操作）。你知道为什么吗？这意味着你有孤儿的记录，在处理这些孤儿之前你无法创建FK。谢谢你对艾哈迈德的精彩解释。但是，我得到了以下两个错误：注意：未定义索引：中的content\u id和警告：mysqli\u fetch\u array（）期望参数1为mysqli\u result，给定布尔值。知道原因吗？select语句中未选择内容id。结果是空的，所以首先必须检查它是否有超过零的记录，然后继续。我已相应地更新了我的答案。现在看看
function get_comment_id($username) {
    include('db_connection.php');
    $username = sanitize($username);
    $sql = "SELECT comment_id FROM `article_comments` WHERE username = '$username'";
    $query = mysqli_query($dbCon, $sql);
    return (mysqli_result($query, 0, 'comment_id'));
}